Question #0cd0c

1 Answer
Jun 17, 2016

Answer:

For a zero-order reaction.

Explanation:

The rate of a reaction is simply a measure of the change in the concentration of the reactants or of the products of a chemical reaction per unit of time.

#"rate" = "change in concentration"/"unit of time"#

For a general #n#-order reaction in which

#overbrace("A" + "A" + ... + "A")^(color(blue)(n)) -> " products"#

you can use the differential rate law to express the relationship between the rate of the reaction, the rate constant #k#, and the concentration of the reactant #["A"]#

#"rate" = ["A"]^color(blue)(n)" " " "color(orange)("(*)")#

The concentration of the reactants is measured in

#["A"] = "mol L"^(-1)#

The rate of the reaction usually expresses the change in concentration of the reactant per second, which means that the units for the rate of the reaction are

#"rate" = "mol L"^(-1) "s"^(-1)#

If you use these units in equation #color(orange)("(*)")#, you will get

#"mol L"^(-1)"s"^(-1) = k * ("mol L"^(-1))^color(blue)(n)#

Rearrange to isolate #k# on one side of the equation

#k = ("mol L"^(-1)"s"^(-1))/("mol L"^(-1))^color(blue)(n)#

You want #k# to have the same units as the rate of the reaction

#k = "mol L"^(-1)"s"^(-1)#

so plug this into the above equation and solve for #color(blue)(n)#

#"mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/("mol L"^(-1))^color(blue)(n)#

As you can see, the only way to have a valid equality here is if #color(blue)(n)=0#, since

#("mol L"^(-1))^0 = 1#

This of course implies that you have

#"mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/1 = "mol L"^(-1)"s"^(-1)" " " "color(green)(sqrt())#

Therefore, the rate constant of a zero-order reaction has the same units as the rate of the reaction, i.e. #"mol L"^(-1)"s"^(-1)#.